Consider the following problem, item 3. of the exercise 1.2 in Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations:
Let $E$ be a vector space of dimension $n$ and let $(e_i)_{1 \leq i \leq n}$ be a basis of $E$. Given $x \in E$ write $x = \sum_1^n x_i e_i$ with $x_i \in \Bbb{R}$. Given $f \in E^*$, define $f_i = \langle f, e_i \rangle$. Consider on $E$ the norm $||x||_p = \left(\sum_1^n |x_i|^p \right)^{1/p}$. (a) Compute $||f||_{E^*}$ explicitly, in terms of the $f_i$. (b) Determine explicitly the duality map for every $x \in E$.
The duality map is defined as follows:
The duality map is the set $F(x) = \{f \in E^* \ : \ ||f|| = ||x|| \text{ and } \langle f, x \rangle = ||x||^2\}$.
I managed to solve part (a). For part (b), the book provides the following solution:
$f \in F(x)$ if and only if $f_i = |x_i|^{p-2}x_i /||x||_p^{p-2}$.
I managed to show that if $f$ is of the form above, then it is in $F(x)$. How to argue for the converse statement?
Thanks in advance.
Best Answer
If $f\in F(x)$, then you have $$\tag1 \left(\sum_j |f_j|^q\right)^{1/q}=\left(\sum_j|x_j|^p\right)^{1/p} $$ and $$\tag2 \sum_j x_j f_j=\left(\sum_j|x_j|^p\right)^{2/p}. $$ Using Hölder, we get from $(2)$ and $(1)$ that $$ \left(\sum_j|x_j|^p\right)^{2/p}=\sum_jx_jf_j\leq\left(\sum_j|x_j|^p\right)^{1/p}\left(\sum_j |f_j|^q\right)^{1/q}=\left(\sum_j|x_j|^p\right)^{2/p}. $$ Thus we have equality in Hölder, which means that $(|f(e_j)|^q)_j=\alpha(|x_j|^p)_j$ for some $\alpha>0$. That is, $$\tag3 |f(e_j)|^q= \alpha |x_j|^p,\ \ j=1,\ldots, n. $$ The equalities also imply that $x_jf_j\geq0$ for all $j$ (because $\sum x_jf_j=\sum|x_jf_j|$). So we can rewrite $(3)$ as $$ f_j=\alpha^{1/q}\,|x_j|^{p/q}\,\operatorname{sign}(x_j)=\alpha^{1/q}\,|x_j|^{p-1}\,\operatorname{sign}(x_j). $$ If we plug this into $(2)$, we get $$ \alpha^{1/q}=\frac{\|x\|_p^2}{\|x\|_p^p}=\|x\|^{2-p} $$ Then $$ f_j=\frac{|x_j|^{p-1}\,\operatorname{sign}(x_j)}{\|x\|_p^{p-2}}=\frac{|x_j|^{p-2}\,x_j}{\|x\|_p^{p-2}}. $$